\newproblem{lay:4_2_3}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.2.3}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Find an explicit description of the null space of $A$ by listing the vectors that span it.
	\begin{center}
		$A=\begin{pmatrix}1 & 2 & 4 & 0 \\ 0 & 1 & 3 & -2\end{pmatrix}$
	\end{center}
}{
  % Solution
	The null space of $A$ is defined as those vectors such that
	\begin{center}
		$A\mathbf{x}=\mathbf{0}$
	\end{center}
	If we construct the augmented matrix of this equation system we get
	\begin{center}
		$\left(\begin{array}{rrrr|r}1 & 2 & 4 & 0 & 0 \\ 0 & 1 & 3 & -2 & 0\end{array}\right) \sim
		 \left(\begin{array}{rrrr|r}1 & 0 & -2 & 4 & 0 \\ 0 & 1 & 3 & -2 & 0\end{array}\right)$
	\end{center}
	So all points satisfying $A\mathbf{x}=\mathbf{0}$ are of the form
	\begin{center}
		$\left.\begin{array}{c}x_1=2x_3-4x_4\\x_2=-3x_3+2x_4\end{array}\right\}\Rightarrow \mathbf{x}=x_3(2,-3,1,0)+x_4(-4,2,0,1)$
	\end{center}
	So a basis of $\mathrm{Nul}\{A\}$ is given by
	\begin{center}
		$\mathrm{Basis}\{\mathrm{Nul}\{A\}\}=\{(2,-3,1,0),(-4,2,0,1)\}$
	\end{center}
}
\useproblem{lay:4_2_3}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
